The minimum possible value of the integral

[jeremjen]

The function has a continuous derivative on the interval [math][-5;5][/math], except, perhaps, for a finite number of points, at each of which the derivative has finite one-sided limits. It is known that [math]f(-5)=f(5)=0[/math] and [math]f(x)≥√5[/math] at [math]x∈[-3;3][/math]. The minimum possible value of the integral [math]\int^5_{-5} \frac {\sqrt{1+f'(x)^2}}{1}dx[/math] is _________.

 

[Deleted member 4993]

The function has a continuous derivative on the interval [math][-5;5][/math], except, perhaps, for a finite number of points, at each of which the derivative has finite one-sided limits. It is known that [math]f(-5)=f(5)=0[/math] and [math]f(x)≥√5[/math] at [math]x∈[-3;3][/math]. The minimum possible value of the integral [math]\int^5_{-5} \frac {\sqrt{1+f'(x)^2}}{1}dx[/math] is _________.

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[Dr.Peterson]

The function has a continuous derivative on the interval [math][-5;5][/math], except, perhaps, for a finite number of points, at each of which the derivative has finite one-sided limits. It is known that [math]f(-5)=f(5)=0[/math] and [math]f(x)≥√5[/math] at [math]x∈[-3;3][/math]. The minimum possible value of the integral [math]\int^5_{-5} \frac {\sqrt{1+f'(x)^2}}{1}dx[/math] is _________.

Do you recognize that the integral represents?

If so, what would it mean to minimize it?

Keep in mind that we're here to help you think, not to let you can get away without thinking at all.

 

[Steven G]

Please draw a graph that depicts what you are given.
Without doing calculus at all, can you state what that integral from x=-3 to x=+6 equals.